# How do you solve #(8x)^(1/2)+6=0#?

##### 2 Answers

Jul 27, 2016

#### Explanation:

Get rid of 6 from left side

For that subtract 6 on both sides

Squaring on both sides

Jul 28, 2016

There are no values of

#### Explanation:

#(8x)^(1/2)+6=0#

Subtract

#(8x)^(1/2) = -6#

Square both sides, noting that squaring may introduce spurious solutions:

#8x = 36#

Divide both sides by

#x = 36/8 = 9/2#

Check:

#(8x)^(1/2)+6 = (8*9/2)^(1/2)+6 = 36^(1/2)+6 = 6+6 = 12#

So this

The problem is that while

So the original equation has no solutions (Real or Complex).